2. I asked the girl to update her lessons.

3. I asked the girl not to learn her lessons parrot-fashion.

4. I asked the girl if she read many books.

5. I asked the girl how many books she bought every year.

nó cứ sai sai sao ý

\(\frac{75.5^4+175.5^4}{20.25.125-625.75}=\frac{\left(75+175\right).5^4}{4.5.25.5^3-5^4.75}\)

\(=\frac{250.5^4}{100.5^4-5^4.75}=\frac{250.5^4}{\left(100-75\right).5^4}\)

\(=\frac{250}{25}=10\)

Câu 4)

Có 3 dạng cơ năng

- thế năng hấp dẫn : quả bính đang bay 

- thế năng đần hồi : lò xo

- động năng : ô tô đang chạy

Câu 5)

Năng lượng vẫn đc bảo toàn và nó chỉ chuyển từ dạng này sang dạng khác

b: Ta có: \(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\dfrac{-64}{125}\)

\(\Leftrightarrow\dfrac{3}{5}-\dfrac{2}{3}x=\dfrac{-4}{5}\)

\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{3}{5}+\dfrac{4}{5}=\dfrac{7}{5}\)

hay \(x=\dfrac{7}{5}:\dfrac{2}{3}=\dfrac{21}{10}\)

Bạn cập nhật lại đề lên nhé

Câu 21 : 

Số chất là 2 ( Fe₂O₃,Mg)

\(Fe_2O_3 + 3H_2SO_4\ to Fe_2(SO_4)_3 + 3H_2O\\ Mg + H_2SO_4 \to MgSO_4 + H_2\\ Mg + 2H_2SO_4 \to MgSO_4 + SO_2 + 2H_2O\)

Câu 20 : 

\(n_{SO_4^{2-}} = n_{H_2SO_4} = n_{H_2} = \dfrac{11,2}{22,4} = 0,5(mol)\\ m_{muối} = m_{kim\ loại} + m_{SO_4^{2-}} = 13,4 + 0,5.96 = 61,4(gam)\)

7. C

8. D

9. A

10. B

31/

\(3z^2-2z+27=0\)

\(\Delta'=\left(-1\right)^2-3.27=1-3.27=-80\)

\(\Delta'\) có 2 căn bậc 2 là \(\pm4i\sqrt{5}\)

\(\Rightarrow\left\{{}\begin{matrix}z_1=\dfrac{1+4i\sqrt{5}}{3}\\z_2=\dfrac{1-4i\sqrt{5}}{3}\end{matrix}\right.\Rightarrow\left|z_1\right|=\left|z_2\right|=\sqrt{\left(\dfrac{1}{3}\right)^2+\left(\dfrac{4\sqrt{5}}{3}\right)^2}=3\)

\(\Rightarrow z_1\left|z_2\right|+z_2\left|z_1\right|=1+4i\sqrt{5}+1-4i\sqrt{5}=2\) => A

32/ \(\Delta'=4-29=-25\Rightarrow\left\{{}\begin{matrix}z_1=-2+5i\\z_2=-2-5i\end{matrix}\right.\Rightarrow\left|z_1\right|=\left|z_2\right|=\sqrt{2^2+5^2}=\sqrt{29}\)

\(\Rightarrow\left|z_1\right|^4+\left|z_2\right|^4=2.\sqrt{29^4}=1682\) => B

33/ \(\Delta=1-12=-11\Rightarrow\left\{{}\begin{matrix}z_1=\dfrac{1+i\sqrt{11}}{6}\\z_2=\dfrac{1-i\sqrt{11}}{6}\end{matrix}\right.\Rightarrow\left|z_1\right|=\left|z_2\right|=\sqrt{\left(\dfrac{1}{6}\right)^2+\left(\dfrac{\sqrt{11}}{6}\right)^2}=\dfrac{\sqrt{3}}{3}\)

\(\Rightarrow\left|z_1\right|+\left|z_2\right|=\dfrac{2\sqrt{3}}{3}\) => D

34/ \(\Delta=1-4.3.2=-23\Rightarrow\left\{{}\begin{matrix}z_1=\dfrac{1-i\sqrt{23}}{6}\\z_2=\dfrac{1+i\sqrt{23}}{6}\end{matrix}\right.\Rightarrow\left|z_1\right|=\left|z_2\right|=\sqrt{\dfrac{1}{36}+\dfrac{23}{36}}=\dfrac{\sqrt{6}}{3}\)

\(\Rightarrow T=2.\left(\dfrac{\sqrt{6}}{3}\right)^2=\dfrac{4}{3}\) => C